Question

How do you convert `r=4/(1-costheta)` to rectangular form?

Answer 1

`y^2 = 8x+16`

Explanation:

We have:

`x = r cos theta`

`y = r sin theta`

`r = sqrt(x^2+y^2)`

Given:

`r = 4/(1-cos theta)`

Multiply both sides by `(1-cos theta)` to get:

`r - r cos theta = 4`

So we have:

`sqrt(x^2+y^2) - x = 4`

We could express this equation in other ways, but note that `sqrt(x^2+y^2)` is the non-negative square root. So if our reexpression involves eliminating the square root by squaring then we need the restriction `x >= -4`.

Add `x` to both sides to get:

`sqrt(x^2+y^2) = x+4`

Square both sides (noting comments above) to get:

`x^2+y^2 = x^2+8x+16`

Subtract `x^2` from both sides to get:

`y^2 = 8x+16 = 8(x+2)`

Now note that `y^2 >= 0` for any Real value of `y`.

Hence `x >= -2` which satisfies the requirement `x >= -4`

So we do not need to explcitly limit the domain and can state:

`y^2 = 8x+16`

graph{y^2 = 8x+16 [-10, 10, -5, 5]}